A band-limited signal has a Fourier component containing a heaviside step function:
<br />
X(f) = c(f) \theta(B - |f|)<br />
The Fourier transform of a product of two functions is their convolution:
<br />
\begin{array}{l}<br />
\mathrm{F.T.}[a(t) b(t)](f) = \int_{-\infty}^{\infty}{a(t) \, b(t) \, e^{-2\pi j f t} \, dt} \\ \\<br />
<br />
= \int_{-\infty}^{\infty}{dt \, e^{-2\pi j f t} \, \int_{-\infty}^{\infty}{A(f_{1}) \, e^{2\pi j f_{1} t} \, df_{1}} \, \int_{-\infty}^{\infty}{B(f_{2}) \, e^{2\pi j f_{2} t} \, df_{2}}} \\ \\<br />
<br />
= \int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{df_{1} \, df_{2} \, A(f_{1}) \, B(f_{2}) \, \int_{-\infty}^{\infty}{dt \, e^{2\pi j (f_{1} + f_{2} - f) t} \, dt}}} \\ \\<br />
<br />
= \int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{df_{1} \, df_{2} \, A(f_{1}) \, B(f_{2}) \, \delta(f_{1} + f_{2} - f)} \\ \\<br />
<br />
= \int_{-\infty}^{\infty}{A(f') \, B(f - f') \, df'}<br />
\end{array}<br />
Take the signals A(f) and B(f) to be band-limited with band limits B_{1} and B_{2}, respectively. Then, their convolution is:
<br />
A \ast B(f) = \int_{-\infty}^{\infty}{A(f') \, \theta(B_{1} - |f'|) \, B(f - f') \, \theta(B_{2} - |f - f'|) \, df'}<br />
The integrand is non-zero only when:
<br />
\left\{\begin{array}{lcl}<br />
B_{1} - |f'| & \ge & 0 \\<br />
<br />
B_{2} - |f - f'| & \ge & 0<br />
\end{array}\right.<br />
These conditions limit the domain of integration with respect to f'. But, they also give some necessary conditions on the possible values of f when the above conditions are not contradictory. This makes the convolution also band limited. What is the band limit on the convolution in terms of B_{1} and B_{2}?
Then, use Mathematical Induction to prove the band limit of the n-th power.
